The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 53 ounces and a standard deviation of 11 ounces. Use the Empirical Rule. Suggestion: sketch the distribution in order to answer these questions. a) $$ 95 \% $$ of the widget weights lie between 31 75 b) What percentage of the widget weights lie between 42 and 75 ounces? c) What percentage of the widget weights lie below 86 ?

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UpStudy AI · Accepted Answer

To solve the problem, we will use the Empirical Rule, which states that for a normal distribution:

1. Approximately $$68\%$$ of the data falls within one standard deviation of the mean.
2. Approximately $$95\%$$ of the data falls within two standard deviations of the mean.
3. Approximately $$99.7\%$$ of the data falls within three standard deviations of the mean.

### Given:
- Mean ($$\mu$$) = 53 ounces
- Standard Deviation ($$\sigma$$) = 11 ounces

### Step-by-step Solution:

#### a) Find the range for $$95\%$$ of the widget weights.

Using the Empirical Rule:
- The range for $$95\%$$ of the data is given by:
  $$
  \mu - 2\sigma \quad \text{to} \quad \mu + 2\sigma
  $$

Calculating the values:
- Lower limit: 
  $$
  53 - 2 \times 11 = 53 - 22 = 31 \text{ ounces}
  $$
- Upper limit: 
  $$
  53 + 2 \times 11 = 53 + 22 = 75 \text{ ounces}
  $$

Thus, $$95\%$$ of the widget weights lie between **31 ounces and 75 ounces**.

#### b) Find the percentage of widget weights between 42 and 75 ounces.

First, we need to determine how many standard deviations away from the mean these values are:

- For 42 ounces:
  $$
  z = \frac{42 - 53}{11} = \frac{-11}{11} = -1
  $$
- For 75 ounces:
  $$
  z = \frac{75 - 53}{11} = \frac{22}{11} = 2
  $$

Using the Empirical Rule:
- The percentage of data between $$z = -1$$ and $$z = 2$$:
  - From $$z = -1$$ to $$z = 1$$ is $$68\%$$ (which includes the mean).
  - From $$z = 1$$ to $$z = 2$$ is half of the remaining $$32\%$$ (since $$95\%$$ is between $$z = -2$$ and $$z = 2$$), which is $$16\%$$.

Thus, the percentage of widget weights between 42 and 75 ounces is:
$$
68\% + 16\% = 84\%
$$

#### c) Find the percentage of widget weights below 86 ounces.

Calculating the z-score for 86 ounces:
$$
z = \frac{86 - 53}{11} = \frac{33}{11} = 3
$$

Using the Empirical Rule:
- Approximately $$99.7\%$$ of the data falls within $$3$$ standard deviations of the mean. Therefore, the percentage of widget weights below $$86$$ ounces is:
$$
99.7\%
$$

### Summary of Results:
- a) $$95\%$$ of the widget weights lie between **31 ounces and 75 ounces**.
- b) The percentage of widget weights between **42 and 75 ounces** is **84%**.
- c) The percentage of widget weights below **86 ounces** is **99.7%**.
- a) 95% of the widget weights are between 31 ounces and 75 ounces. - b) 84% of the widget weights are between 42 and 75 ounces. - c) 99.7% of the widget weights are below 86 ounces.

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